LeetCode每日一题——2525. Categorize Box According to Criteria

2023-10-20 11:04

本文主要是介绍LeetCode每日一题——2525. Categorize Box According to Criteria,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!

文章目录

    • 一、题目
    • 二、题解

一、题目

Given four integers length, width, height, and mass, representing the dimensions and mass of a box, respectively, return a string representing the category of the box.

The box is “Bulky” if:
Any of the dimensions of the box is greater or equal to 104.
Or, the volume of the box is greater or equal to 109.
If the mass of the box is greater or equal to 100, it is “Heavy”.
If the box is both “Bulky” and “Heavy”, then its category is “Both”.
If the box is neither “Bulky” nor “Heavy”, then its category is “Neither”.
If the box is “Bulky” but not “Heavy”, then its category is “Bulky”.
If the box is “Heavy” but not “Bulky”, then its category is “Heavy”.
Note that the volume of the box is the product of its length, width and height.

Example 1:

Input: length = 1000, width = 35, height = 700, mass = 300
Output: “Heavy”
Explanation:
None of the dimensions of the box is greater or equal to 104.
Its volume = 24500000 <= 109. So it cannot be categorized as “Bulky”.
However mass >= 100, so the box is “Heavy”.
Since the box is not “Bulky” but “Heavy”, we return “Heavy”.
Example 2:

Input: length = 200, width = 50, height = 800, mass = 50
Output: “Neither”
Explanation:
None of the dimensions of the box is greater or equal to 104.
Its volume = 8 * 106 <= 109. So it cannot be categorized as “Bulky”.
Its mass is also less than 100, so it cannot be categorized as “Heavy” either.
Since its neither of the two above categories, we return “Neither”.

Constraints:

1 <= length, width, height <= 105
1 <= mass <= 103

二、题解

class Solution {
public:string categorizeBox(int length, int width, int height, int mass) {bool isBulky = false,isHeavy = false;long long V = (long long)length * width * height;if(length >= pow(10,4) || width >= pow(10,4) || height >= pow(10,4) || V >= pow(10,9)) isBulky = true;if(mass >= 100) isHeavy = true;if(isBulky && isHeavy) return "Both";else if(!isBulky && !isHeavy) return "Neither";else if(isBulky && !isHeavy) return "Bulky";else return "Heavy";}
};

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